How do you evaluate # e^( ( 7 pi)/6 i) - e^( ( 19 pi)/8 i)# using trigonometric functions?

1 Answer
Rajat Chatterjee
Nov 4, 2016

#2sin((29pi)/24)e^(((25pi)/24)i)#

Explanation:

we know euler's theorem states that #e^(itheta)=costheta+isintheta#
so,#e^(7pi/6i)-e^(19pi/8 i)#
=#cos(7pi/6)+isin(7pi/6)-(cos(19pi/8)+isin(19pi/8))#
=#cos(7pi/6)-cos(19pi/8)+i(sin(7pi/6)-sin(19pi/8))#
now #cosc-cosd=2sin((c+d)/2) sin((d-c)/2)#
#sinc-sind=2cos((c+d)/2) sin((c-d)/2)#
so we get #cos((7pi)/6)-cos((19pi)/8)#
=#2sin((85pi)/24)sin((29pi)/24)#
again similarly we get #i(sin((7pi)/6)-sin((19pi)/8))#
=#2icos((85pi)/24)sin((29pi)/24)#
so,#e^((7pi)/6)-e^((19pi)/8)#=#2sin((29pi)/24)(sin((85pi)/24)-icos((85pi)/24))#
=#2sin((29pi)/24)[sin(2pi+((37pi)/24))-icos(2pi+((37pi)/24))]#
=#2sin((29pi)/24)[sin((37pi)/24)-icos((37pi)/24)]#
=#2sin((29pi)/24)[cos((25pi)/24)+isin((25pi)/24)]#
=#2sin((29pi)/24)e^(((25pi)/24)i)#.

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